The sum of all distinct roots of the equation | vedclass

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(B) Given equation is $x^5-3x^4+5x^3-5x^2+3x-1=0$. We can rewrite the equation by grouping terms: $(x^5-1) - 3x(x^3-1) + 5x^2(x-1) = 0$. Factoring out

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$2$

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(B) Given equation is $x^5-3x^4+5x^3-5x^2+3x-1=0$. We can rewrite the equation by grouping terms: $(x^5-1) - 3x(x^3-1) + 5x^2(x-1) = 0$. Factoring out $(x-1)$: $(x-1)(x^4+x^3+x^2+x+1) - 3x(x-1)(x^2+x+1) + 5x^2(x-1) = 0$. $(x-1)[(x^4+x^3+x^2+x+1) - 3x(x^2+x+1) + 5x^2] = 0$. $(x-1)[x^4+x^3+x^2+x+1 - 3x^3-3x^2-3x + 5x^2] = 0$. $(x-1)(x^4-2x^3+3x^2-2x+1) = 0$. For the quartic part,divide by $x^2$ (since $x=0$ is not a root): $x^2-2x+3-\frac{2}{x}+\frac{1}{x^2} = 0$. $(x^2+\frac{1}{x^2}) - 2(x+\frac{1}{x}) + 3 = 0$. Let $y = x+\frac{1}{x}$,then $y^2-2 - 2y + 3 = 0$,so $y^2-2y+1 = 0$. $(y-1)^2 = 0$,which gives $y=1$. $x+\frac{1}{x} = 1 \implies x^2-x+1 = 0$. The roots of $x^2-x+1=0$ are complex. The only real root is $x=1$. The distinct roots are $1, \frac{1+i\sqrt{3}}{2}, \frac{1-i\sqrt{3}}{2}$. The sum of these distinct roots is $1 + (\frac{1+i\sqrt{3}}{2} + \frac{1-i\sqrt{3}}{2}) = 1 + 1 = 2$.

The sum of all distinct roots of the equation $x^5-3x^4+5x^3-5x^2+3x-1=0$ is

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